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Math and science::Analysis::Tao::06. Limits of sequences

# Limits of sequences

If a sequence of reals $$(a_n)_{n=m}^{\infty}$$ converges to a real $$L$$, we say that $$(a_n)_{n=m}^{\infty}$$ is convergent and that $$L$$ is the limit of the sequence. We write:
$L = \lim_{n\to\infty}a_n$

If a sequence does not converge to a real, then it is said to diverge or be divergent. $$\lim_{n\to\infty}a_n$$ is left undefined for a divergent sequence.

$$L$$ is unique as per the Uniqueness of Convergence proposition. The uniqueness allows us to create the limit notation $$\lim_{n\to\infty} a_n$$, dependent only on the sequence $$(a_n)_{n=m}^{\infty}$$, and for that notation to be equated to a single real. If fact, the limit notation doesn't depend on the full specification of the sequence $$(a_n)_{n=m}^{\infty}$$, as the starting index is irrelevant and neglected. So the $$\lim_{n\to\infty} a_n$$ is the limit for any sequences $$(a_n)_{n=N}^{\infty}$$ where $$N$$ is some integer.

### Example

#### $$\lim_{n\to\infty} \frac{1}{n} = 0$$, proof

$$\lim_{n\to\infty} \frac{1}{n}= 0$$ is true if the sequence $$(a_n)_{n=1}^{\infty}$$ converges to 0, where $$a_n = \frac{1}{n}$$.

Let $$\varepsilon > 0$$ be an arbitary real. We must show that there exists an integer $$N$$ such that $$|a_n - 0| \le \varepsilon$$ for all $$n \ge N$$.

$$|a_n - 0| = |\frac{1}{n} - 0| = \frac{1}{n} \le \frac{1}{N}$$.

$$\frac{1}{N} \le \varepsilon$$ is true if $$N \ge \frac{1}{\varepsilon}$$. Such an integer $$N$$ exists as per the Achimedean principal. Thus $$|a_n - 0| \le \varepsilon$$ for all $$n \ge N$$. Thus, $$(a_n)_{n=1}^{\infty}$$ is eventually $$\varepsilon$$ close to 0 for arbitary $$\varepsilon > 0$$. Thus, $$(a_n)_{n=1}^{\infty}$$ converges to 0. This allows us to write $$\lim_{n\to\infty} \frac{1}{n}= 0$$.

Cauchy sequences of reals → convergence of sequences of reals → uniqueness of convergence → limit, the definition → subsumption of formal limits

Tao, Analysis I
Chapter 6